Hello,

I want to show that this function is smooth:

$\displaystyle

f:T->\mathbb{R} , f(x)=e^1(x)=x^1

$

T is the Torus defined as: $\displaystyle T=\{((R+r*cosv)cosu,(R+r*cosv)sinu,r*sinv) \subset \mathbb{R}^3: u,v \in [0,2\pi)\}$ and R,r>0 constant.

I don't know how f could be differentiated. My Problem is, that f is not in the standard form f(x,y,z)=x. it is something like f((R+r.....))=(R+r*cosu)cosv.

I'm a little bit confused. Is here any trick or something

Thank you for your help

regards